# How do you find vertical, horizontal and oblique asymptotes for (3x^2) /( x^2 - 9)?

Apr 19, 2016

vertical asymptotes x = ± 3
horizontal asymptote y = 3

#### Explanation:

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation/s set the denominator equal to zero.

solve : x^2 - 9 =0 → (x-3)(x+3) = 0 → x = ± 3

$\Rightarrow x = - 3 , x = 3 \text{ are the asymptotes }$

Horizontal asymptotes occur as ${\lim}_{x \to \pm \infty} f \left(x\right) \to 0$

divide terms on numerator/denominator by ${x}^{2}$

$\frac{3 {x}^{2}}{x} ^ \frac{2}{{x}^{2} / {x}^{2} - \frac{9}{x} ^ 2} = \frac{3}{1 - \frac{9}{x} ^ 2}$

as $x \to \pm \infty , \frac{9}{x} ^ 2 \to 0 \text{ and } y \to \frac{3}{1}$

$\Rightarrow y = 3 \text{ is the asymptote }$

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here hence there are no oblique asymptotes.
graph{(3x^2)/(x^2-9) [-10, 10, -5, 5]}